You have to be careful when $n=0$. Also your assertion that "the sum is the same since $n\to\infty$" is incorrect - it does converge similarly, but not to the same value...
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gt6989bNov 29 '12 at 18:03

@gt6989b Could you elaborate in an answer how I could alleviate those problems?
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FlaviusNov 29 '12 at 18:11

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@Flavius The first term in the given sum is $0$. So you may omit it and save yourself some $0/0$ grief by noting that your sum equals $\sum_{n=1}^{\infty}\frac{3n}{n!}$. Then reindex by replacing all "n"s by "n+1": $\sum_{n+1=1}^{n+1=\infty}\frac{3}{(n+1-1)!}$.
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alex.jordanNov 29 '12 at 18:18

4 Answers
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Note that the term corresponding to $n=0$ is $0$. This is because $0!$, by definition, is equal to $1$. So, as pointed out by Phira, our sum is equal to $\sum_{n=1}^\infty \frac{3n}{n!}$. As in your answer, this simplifies to $\sum_{n=1}^\infty \frac{3}{n-1}!$. Writing $k$ for $n-1$, we see that our sum is $\sum_{k=0}^\infty \frac{3}{k!}$, which we recognize.

Conceivably, this is not quite the desired answer! Perhaps the problem setter expects us to investigate the series for convergence while pretending we don't know the sum, and as a second part, to find the sum. If so, the Ratio Test quickly tells us that the series converges.

Even though you end up with the right number, the first line of your answer is problematic. For in the case $n=0$, you are cancelling $0$'s (in general forbidden). Then your first term is $\frac{3}{(-1)!}$, whatever that may mean. Then for no clear reason the $(n-1)!$ gets replaced by $n!$.

Beside the point that @Phira noted: $$\sum_{n=0}^\infty\frac{3n}{(n−1)!n}\to\sum_{n=1}^\infty\frac{3n}{(n−1)!n}$$ I think, since the partial sums of the latter series is the same patial sums for $\exp(1)$, then your claim looks right.